# What is the missing proportionality constant in the magnetic levitation formula?

The formula for magnetic levitation is

$$B \frac{dB}{dz} = \frac{ \rho g }{\chi}$$

but as always, I have a hard time figuring the units in SI. The left hand side is $\mathrm{T^2 /m}$, while $\chi$ has units of $\mathrm{m^3\,mol^{-1}}$, which makes the right hand side with units of..

$$\mathrm{kg\,m^{-5}\,s^{-2}\,(mol)}$$

Obviously there must be some constant, but I don't know what is it, as all texts use this natural unit system. Help?

Update:

using the permeability of free space as reported by Wikipedia, which seems to be Tesla-meter per Ampere, leaves the right hand side as

$$\mathrm{kg\,T\,m^{-4}\,s^{-2}\,Ampere\,mol}$$

still far from being recognizable to the left side

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sounds to me like those susceptibility units are somehow messed up: en.wikipedia.org/wiki/… – lurscher Aug 31 '12 at 20:07

As lurscher mentioned in a comment, you're using the wrong units for magnetic susceptibility. $\chi$ is actually a dimensionless number that is related to the magnetic permeability of a material relative to that of a vacuum. I think you were mixing it up with the molar magnetic susceptibility, which is $\chi_\text{mol} = \mathcal{M}\chi/\rho$, where $\mathcal{M}$ is the molar mass of the substance (units of $\mathrm{kg/mol}$) and $\rho$ is the density (units of $\mathrm{kg/m^3}$). $\chi_\text{mol}$ is the thing with units of $\mathrm{m^3/mol}$, but it's $\chi$ that actually appears in the magnetic levitation formula.

With that cleared up, let's look at the equation. The left side, naturally, has units of $\mathrm{T^2/m}$. If you include the magnetic constant on the right side, as Wikipedia (correctly) does, you have

$$\biggl[\mu_0\frac{ \rho g }{\chi}\biggr] = [\mu_0]\frac{ [\rho] [g] }{[\chi]} = \biggl(\frac{\mathrm{T\,m}}{\mathrm{A}}\Biggr)\frac{\mathrm{(kg/m^3)(m/s^2)}}{1} = \frac{\mathrm{T\,kg}}{\mathrm{m\,s^2\,A}}$$

Here I'm using the notation where putting brackets around a quantity designates the units of that quantity. For example, the units of the magnetic constant are $\mathrm{T\,m/A}$, so $[\mu_0] = \mathrm{T\,m/A}$.

Now you can equate the units of the two sides of the equation:

$$\frac{\mathrm{T^2}}{\mathrm{m}} = \frac{\mathrm{T\,kg}}{\mathrm{m\,s^2\,A}}$$

which simplifies to

$$\mathrm{T} = \frac{\mathrm{kg}}{\mathrm{s^2\,A}}$$

So if this equivalence is correct, then it shows that the original equation is dimensionally consistent. And if you look on the Wikipedia page for the Tesla, it does indeed give $\mathrm{T} = \frac{\mathrm{kg}}{\mathrm{s^2\,A}}$ as one of the definitions of that unit.

Alternatively, you could check it using a formula involving magnetic field and current, such as $\vec{F} = I\mathrm{d}\vec{l}\times\vec{B}$. The units of this are $\mathrm{N = A\,m\times T}$, and since $\mathrm{N} = \mathrm{kg\,m/s^2}$, you can set $\mathrm{kg\,m/s^2 = A\,m\,T}$ and find exactly that $\mathrm{T} = \frac{\mathrm{kg}}{\mathrm{s^2\,A}}$. This is a useful trick that keeps you from having to memorize the definitions of all the SI (or other) units.

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ah i see, i blame this doc: www-d0.fnal.gov/hardware/cal/lvps_info/engineering/… so it seems i just need to multiply this number by density and dividing by molar mass. thanks! – diffeomorphism Aug 31 '12 at 22:11

This is something that took me a long time to wrap my head around as well, but there actually doesn't need to be a constant (depending on the system of measurement). You mention SI, which is definitely a preferred system, but I'm going to bring up cgs because it brings some of the underlying funny business to the surface (particularly because of the definitions of the statcoulomb), and at its roots it's not that far from SI.

You break down the units in SI for the LHS and RHS almost correctly, except this formula uses a dimensionless $\chi$, so I start with:

$\frac{T^2}{m} = \frac{kg\cdot}{m^2\cdot s^2}$

I'll rewrite this in cgs units. $T$ is instead $gauss$, $m$ is replaced by $cm$, $kg$ becomes $g$, and $s$ is unchanged. So I get:

$\frac{gauss^2}{cm} = \frac{g}{cm^2\cdot s^2}$

Gauss is a derived unit which, in more "fundamental" units, is $gauss = \frac{g\cdot cm}{statC\cdot s^2}$. I'm still not actually comfortable trying to explain all the intricacies of the statcoulomb (statC); you'll have to look into that a bit more yourself I'm afraid. It ends up being a unit of charge, but comes with a lot of other implications. In any case, putting in that definition of $gauss$ gives:

$\frac{g^2\cdot cm}{statC^2\cdot s^4} = \frac{g}{cm^2\cdot s^2}$

Or, rearranging and simplifying a bit,

$statC = \frac{g^{1/2}\cdot cm^{3/2}}{s}$

That's actually the definition of the statC, so the equation is dimensionally consistent.

As far as I know the Coulomb from SI doesn't behave at all this way, and I don't have a good way of showing dimensional consistency in SI, but I'd be very interested if someone else did.

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$\mu_0$ is missing on the right hand side, and in fact $\chi$ is dimensionless, it is just the ratio between $M$ and $H$. So

$$B \frac{dB}{dz} = \mu_0 \frac{ \rho g }{\chi}$$

The reason:

• the unit of $B^2/\mu_0$ is energy-density ($J/m^3$)
• the unit of $[\rho g]$ is ($N/m^3$)
• as $J=Nm$ everything is fine
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